Question

solve 1/(x^2+1) * y' + xy = 3 for y(0)=0

So I needed to find the integrating factor which is r(x)=e^(integral(p(x)dx))
so here p(x) is just x therefore r(x)=e^(x^2/2)?

Then I am supposed to be able to put it in the form

d/dx(r(x)y)=RHS but this doesn't work because if you take the derivative of that you're supposed to get the LHS. This doesn't work out obviously.

What am I doing wrong? Do I have to break the 1/x^2+1 into partial fraction decomp? But how would that help?

Thanks!

Answer 1

first you need to get y' by itself.

y' + x(x^2+1) y = 3(x^2+1)
Now the integrating factor is e^(1/2 (x^2+1)^2)

try that.

Answer 2

Note that in a linear differential equation, the y' term is not supposed to have any coefficient. Hence, you must turn the differential equation into this form.

[1/(x^2+1) * y'] + xy = 3

Multiply the whole equation by (x^2 + 1)

y' + [x*(x^2 + 1)]*y = 3(x^2 + 1)

Now, p(x) = [x*(x^2 + 1)]
= (x^3 + x)

Integrating factor = I.F. = e^(integral(p(x)dx))
= e^(integral((x^3 + x)dx))

Then, as is true for linear differential equations,

y*(I.F.) = integral(I.F. * q(x))

Where q(x) = 3(x^2 + 1)

Answer 3

extra credit:
why did Arora and I come up with different integrating factors?
Can they both be correct?